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I'm looking for a direct proof of the following identity: $$\frac{1}{\pi}\int_1^x\frac{dt}{t\sqrt{t-1}}\arcsin\left(\sqrt{\frac{x-t}{y-t}}\right)=\arcsin\sqrt{\frac{x}{y}}-\arcsin\sqrt{\frac{1}{y}},$$ for $1<x<y$.

I know this holds because the integral above is a related to some probability (concerning brownian motion) that can be evaluated by some other means.

Any idea will be welcome!

MassiveJack
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  • Maybe you missed an additive constant? I looked at the lhs and rhs numerically, and unless I did something very wrong, the differ by a constant that is approximately $0.261799$. I'll have another look after my cup of tea... – mickep Mar 31 '15 at 08:15
  • @mickep : Well, I hope not! Indeed, for $y=x$, the integral becomes$$\int_1^x\frac{dt}{2t\sqrt{t-1}}=\arcsin(1)-\arcsin(\sqrt{1/x}),$$ as expected, since $1/(2t\sqrt{t-1})$ is the derivative of $-\arcsin(\sqrt{1/t})$. – MassiveJack Mar 31 '15 at 08:36
  • Never mind, I had forgotten a $\sqrt{}$ term in the last arcsin, they seem to coincide. – mickep Mar 31 '15 at 08:36

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One way out is probably to differentiate both sides, and try to show they are equal.

Indeed, we think of fixed $y$ and note that the equality is true for $x=1$. Next, the derivative of the left-hand side is (one term disappear if one uses the general formula) $$ \frac{1}{2\pi\sqrt{y-x\vphantom{)}}}\int_1^x \frac{1}{t\sqrt{(t-1)(x-t)}}\,dt $$ and the derivative of the right-hand side is $$ \frac{1}{2\sqrt{x(y-x)}}. $$ It thus remains to prove that $$ \int_1^x \frac{1}{t\sqrt{(t-1)(x-t)}}\,dt = \frac{\pi}{\sqrt{x}}. $$ Can you proceed from here?

Update Since the integral in the left-hand side is a bit difficult to handle, I give some further steps:

Do the change of variable $t=(x-1)u+1$ (this sends $t=1$ to $u=0$ and $t=x$ to $u=1$). You will (hopefully) get the integral $$ \int_0^1 \frac{1}{\sqrt{u(1-u)}(1+(x-1)u)}\,du $$ Next, do the change of variables $u=\frac{s^2}{1+s^2}$. This will send $u=0$ to $s=0$ and $u=1$ to $s=+\infty$, and the integral becomes $$ \int_0^{+\infty}\frac{2}{1+s^2 x}\,ds. $$ I'm sure you can continue from here.

mickep
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  • yes! It works : the change of variables $u=\sqrt{\frac{x-t}{t-1}}$ leads to the integral $$\int_0^{\infty}\frac{du}{x+u^2}$$ which gives the result. – MassiveJack Mar 31 '15 at 09:26
  • Haha, that is of course simpler than what I did :) Nevertheless, I wish you luck with the Brownian motion. – mickep Mar 31 '15 at 09:28
  • In fact, I did this calculation yesterday for some related reasons after having heard of Euler substition trick http://en.wikipedia.org/wiki/Euler_substitution. Thank you for your help! – MassiveJack Mar 31 '15 at 09:33